OCP-1Z0-051-V9.02-57题

57. The CUSTOMERS table has the following structure:
name                     Null         Type
CUST_ID    NOT NULL  NUMBER
CUST_FIRST_NAME   NOT NULL  VARCHAR2(20)
CUST_LAST_NAME   NOT NULL  VARCHAR2(30)
CUST_INCOME_LEVEL     VARCHAR2(30)
CUST_CREDIT_LIMIT     NUMBER
You need to write a query that  does the following tasks:
1. Display the first name and tax amount of the customers. Tax is 5% of their credit limit. 
2. Only those customers whose income level has a value should be considered.
3. Customers whose tax amount is null should not be considered.
Which statement accomplishes all the required tasks?
A. SELECT cust_first_name, cust_credit_limit * .05 AS TAX_AMOUNT    
FROM customers    
WHERE cust_income_level IS NOT NULL AND          
tax_amount IS NOT NULL; 
B. SELECT cust_first_name, cust_credit_limit * .05 AS TAX_AMOUNT    
FROM customers    
WHERE cust_income_level IS NOT NULL AND          
cust_credit_limit IS NOT NULL;
C. SELECT cust_first_name, cust_credit_limit * .05 AS TAX_AMOUNT    
FROM customers    
WHERE cust_income_level <> NULL AND          
tax_amount <> NULL;
D. SELECT cust_first_name, cust_credit_limit * .05 AS TAX_AMOUNT    
FROM customers    
WHERE (cust_income_level,tax_amount) IS NOT NULL;
Answer: B
答案解析:
A,WHERE子句不能跟别名
sh@TEST0924> SELECT cust_first_name, cust_credit_limit * .05 AS TAX_AMOUNT FROM customers
  2  WHERE cust_income_level IS NOT NULL AND TAX_AMOUNT IS NOT NULL
  3  /
WHERE cust_income_level IS NOT NULL AND TAX_AMOUNT IS NOT NULL
                                        *
ERROR at line 2:
ORA-00904: "TAX_AMOUNT": invalid identifier

sh@TEST0924> SELECT cust_first_name, cust_credit_limit * .05 AS TAX_AMOUNT FROM customers
  2  WHERE cust_income_level IS NOT NULL AND cust_credit_limit * .05 IS NOT NULL;


CUST_FIRST_NAME      TAX_AMOUNT
-------------------- ----------
Abigail                      75
Abigail                     350
...
B,正确
sh@TEST0924> SELECT cust_first_name, cust_credit_limit * .05 AS TAX_AMOUNT FROM customers
  2  WHERE cust_income_level IS NOT NULL AND cust_credit_limit IS NOT NULL;


CUST_FIRST_NAME      TAX_AMOUNT
-------------------- ----------
Abigail                      75
Abigail                     350
...
C错误,一是where子句不能用别名,二是非null不能用<>来表达
sh@TEST0924> SELECT cust_first_name, cust_credit_limit * .05 AS TAX_AMOUNT FROM customers
  2  WHERE cust_income_level <> NULL AND tax_amount <> NULL;
WHERE cust_income_level <> NULL AND tax_amount <> NULL
                                    *
ERROR at line 2:
ORA-00904: "TAX_AMOUNT": invalid identifier


sh@TEST0924> SELECT cust_first_name, cust_credit_limit * .05 AS TAX_AMOUNT FROM customers
  2  WHERE cust_income_level <> NULL AND cust_credit_limit<> NULL;

no rows selected
D错误,语法错误
sh@TEST0924> SELECT cust_first_name, cust_credit_limit * .05 AS TAX_AMOUNT FROM customers
  2  WHERE (cust_income_level,tax_amount) IS NOT NULL;
WHERE (cust_income_level,tax_amount) IS NOT NULL
                                     *
ERROR at line 2:
ORA-00920: invalid relational operator
原文地址:https://www.cnblogs.com/hzcya1995/p/13316913.html